Power Series of $\ \frac x3 \ln(1+x^2).$ I have a question regarding the power series representation of the function $$\ \frac x3 \ln(1+x^2).$$ 
I got the answer $$\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{2n+1}}}{3n} $$
by using the Maclaurin series expansion for $\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{n}}}{n}$ and replacing $x$ with $x^2$ and multiplying the series with $\frac{x}{3}$. 
However, my teacher got the answer $$\frac 23\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{2n+3}}}{2n+1}. $$
My teacher mentioned differentiation and integration, as one of the first steps in his method was taking $\ln(1+x^2)$ and representing it as $\int \frac{2x dx}{1+x^2}.$ He then represented the original function as $\frac x3 \int \frac{2x dx}{1+x^2}.$ However, after this I could not follow the rest of the procedures. I am completely lost as to how my teacher obtained this answer, and any help would be appreciated in explaining how one may arrive to this conclusion.
 A: So continuing from where your teacher left off, we have $\frac{2x}3 \int\frac{xdx}{1+x^2}$. We start with the series expansion for $\frac{1}{1+y}$:
$$ \frac1{1+y} = \sum_{n=0}^\infty (-1)^ny^n$$
We substitute $y = x^2$:
$$ \frac1{1+x^2} = \sum_{n=0}^\infty (-1)^nx^{2n}$$
We multiply by $x$:
$$ \frac x{1+x^2} = \sum_{n=0}^\infty (-1)^nx^{2n+1}$$
And now we integrate:
$$\int\frac{xdx}{1+x^2} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n+2}}{2n+2} $$
Finally, we multiply by $\frac{2x}3$:
$$ \frac{2x}3 \int\frac{xdx}{1+x^2} = \frac23 \sum_{n=0}^\infty \frac{(-1)^nx^{2n+3}}{2n+2}$$
I imagine this was the intended methodology, but the answer I get doesn't match your teacher's. In fact, it matches yours by making the change of index $m=n+1$:
$$\frac{2x}3 \int\frac{xdx}{1+x^2} = \frac23 \sum_{m=1}^\infty \frac{(-1)^{m-1}x^{2m+1}}{2m}= \frac13 \sum_{m=1}^\infty \frac{(-1)^{m+1}x^{2m+1}}{m}$$
A: Also,
Wolfy says that your teacher's sum is
$\sum_{n=1}^∞ \dfrac{(-1)^{n + 1} x^{2 n + 3}}{2 n + 1} 
= -x^2 (\tan^{-1}(x) - x)
$
